Question

Find all the zeroes of 2x^4 - 3x^3 - 3x^2 + 6x - 2, the two of zeroes are √2 and -√2.

Answer 1

Question :–

Find all the zeroes of ${2x}^{4} - {3x}^{3} - {3x}^{2} + 6x - 2,$ the two of zeroes are √2 and -√2.

Solution :–

Since √2 and -√2 are zeroes of p(x).

So, (x - √2) and (x + √2) will divide p(x).

i.e. (x - √2) and (x + √2) will be divide p(x).

→ (x² - 2) will be also divide p(x).

→ p(x) = ${2x}^{4} - {3x}^{3} - {3x}^{2} + 6x - 2,$

★ Division is refer to attachment ★

q(x) = 2x² - 3x + 1

= 2x² - 2x - 1x + 1

= 2x (x - 1) - 1 (x - 1)

= (x - 1) (2x - 1)

So, other two zeroes :–

⓵ x - 1 = 0

= x = 1

⓶ 2x - 1 = 0

= 0x = $\sf \dfrac{1}{2}$

Answer 2

Answer:

Step-by-step explanation:

q(x) = 2x² - 3x + 1

= 2x² - 2x - 1x + 1

= 2x (x - 1) - 1 (x - 1)

= (x - 1) (2x - 1)

So, other two zeroes :–

⓵ x - 1 = 0

= x = 1

⓶ 2x - 1 = 0

= 0x =  \sf \dfrac{1}{2}